Theorem ddif 4089

Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.)

Assertion

Ref	Expression
ddif	⊢ (V ∖ (V ∖ 𝐴)) = 𝐴

Proof of Theorem ddif

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		vex 3438	. . . . 5 ⊢ 𝑥 ∈ V
2		eldif 3910	. . . . 5 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ 𝐴))
3	1, 2	mpbiran 709	. . . 4 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥 ∈ 𝐴)
4	3	con2bii 357	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ ¬ 𝑥 ∈ (V ∖ 𝐴))
5	1	biantrur 530	. . 3 ⊢ (¬ 𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)))
6	4, 5	bitr2i 276	. 2 ⊢ ((𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)) ↔ 𝑥 ∈ 𝐴)
7	6	difeqri 4076	1 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴

Syntax hints:  ¬ wn 3   ∧ wa 395   = wceq 1541   ∈ wcel 2110  Vcvv 3434   ∖ cdif 3897

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2112  ax-9 2120  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1544  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2722  df-clel 2804  df-v 3436  df-dif 3903

This theorem is referenced by:  complss  4099  dfun3  4224  dfin3  4225  invdif  4227  ssindif0  4412  difdifdir  4440